Figure (8-E4) sho

Let the mass of the particle be m

Acceleration due to gravity, g = 9.8 m/s²

The height of the particle at point A, h = 1.0 m

Height of particle at the point where it terminates into straight horizontal section, h= 0.5 m

Applying energy conservation on the particle, total initial energy = total final energy

⇒ (Kinetic energy)_initial + (Potential energy)_initial = (Kinetic energy)_final + (Potential energy)_final

⇒ 0 + mgh_{at A} = 1/2 mv² + mgh_{at point of termination}

⇒ m×9.8×1 = 0.5×m×v² + m×9.8×0.5

⇒ 9.8 = 0.5v² + 4.9

⇒ v² = 9.8

⇒ v = 3.13 m/s

Let the time taken by particle to reach the termination point be t.

The particle starts sliding from rest, from the equation of motion

⇒ h = 1/2 gt²

⇒ 0.5 = 0.5 × 9.8 × t²

⇒ t² = 0.1020

⇒ t = 0.32 s

∴ The horizontal distance that the particle will travel = speed v × time t

= 3.13 × 0.32

= 1.0016 ≈ 1m